已反编译的venderInfo的子类：
import com.macrovision.flexlm.misc.FlexlmPublicKey;
import java.security.PublicKey;
import com.macrovision.flexlm.FlexlmException;
import com.macrovision.flexlm.FlexlmConstants;
import com.macrovision.flexlm.VendorInfo;
public class XXXXInfo extends VendorInfo implements FlexlmConstants
{
public XXXXInfo() throws FlexlmException {
super();
}
public String getVendorName() {
return "略去 ";
}
public PublicKey getPublicKey(final int n) {
switch (n) {
case 2: {
return (PublicKey)new FlexlmPublicKey(this.getVendorName(), "659792E05C86A46B0B4AB31F10A53B02");
}
case 3: {
return (PublicKey)new FlexlmPublicKey(this.getVendorName(), "66994BE6728C8304C0A951B450303C6DE523F0A4C38C");
}
case 4: {
return (PublicKey)new FlexlmPublicKey(this.getVendorName(), "66BEF0D5F921DC63DC67FAD8873C6EA4D887AB21C5A99D6BB56A792CFA8A15");
}
default: {
return null;
}
}
}
public int[] getEncryptionSeeds() {
return new int[] { 241158379, 1268573936 };
}
public int[] getVendorKeys() {
return new int[] { -1693521973, -1789841632, 683175751, -210946792 };
}
public int[] getCroKeys() {
return new int[] { 1611626693, -1841157379 };
}
public int getDefaultStrength() {
return 4;
}
}
下载了flexlm 10.8 Java 版和 flexlm 10.8 版后，经反编译 flexlmutil.jar 发现: Java 版的vendor key 只有4 个, 而且Java的EncryptionSeeds和Native C 版有一下关联:
protected void generateGetEncryptionSeeds(PrintWriter paramPrintWriter)
{
paramPrintWriter.println();
paramPrintWriter.println("\tpublic int[] getEncryptionSeeds() {");
paramPrintWriter.print("\t\treturn new int[] {");
long l1 = parseHexItem("ENCRYPTION_SEED1");
long l2 = parseHexItem("ENCRYPTION_SEED2");
long l3 = parseHexItem("VENDOR_KEY5");
if (this.includeNonCroSeeds)
{
paramPrintWriter.print("0x" + Long.toString(l1 ^ l3, 16));
paramPrintWriter.print(",");
paramPrintWriter.print("0x" + Long.toString(l2 ^ l3, 16));
}
else
{
long l4 = parseHexItem("VENDOR_KEY1");
long l5 = parseHexItem("VENDOR_KEY2");
long l6 = l4 ^ l5;
paramPrintWriter.print("0x" + Long.toString(l6, 16));
paramPrintWriter.print(",");
long l7 = l1 ^ l2 ^ l6;
paramPrintWriter.print("0x" + Long.toString(l7, 16)); //Native 版的EncryptionSeeds 经过异或运算后，被隐藏了
}
paramPrintWriter.println(" };");
paramPrintWriter.println("\t}");
}
protected void generateGetVendorKeys(final PrintWriter printWriter) {
printWriter.println();
printWriter.println("\tpublic int[] getVendorKeys() {");
printWriter.print("\t\treturn new int[] {");
printWriter.print(this.getVendorWord("VENDOR_KEY1")); // Vendor key 1-4 都是从Native C 复制过来的
printWriter.print(",");
printWriter.print(this.getVendorWord("VENDOR_KEY2"));
printWriter.print(",");
printWriter.print(this.getVendorWord("VENDOR_KEY3"));
printWriter.print(",");
printWriter.print(this.getVendorWord("VENDOR_KEY4"));
printWriter.println(" };");
printWriter.println("\t}");
}
protected void generateGetCroKeys(final PrintWriter printWriter) {
printWriter.println();
printWriter.println("\tpublic int[] getCroKeys() {");
printWriter.print("\t\treturn new int[] {");
final long hexItem = this.parseHexItem("TRL_KEY1"); //CroKey 也是从Native C 复制过来的
final long hexItem2 = this.parseHexItem("TRL_KEY2");
if (hexItem != 0L && this.publicKeys != null) {
printWriter.print("0x" + Long.toString(hexItem, 16));
printWriter.print(",");
printWriter.print("0x" + Long.toString(hexItem2, 16));
}
printWriter.println(" };");
printWriter.println("\t}");
}
从代码上看不到逆向求解Native C EncryptionSeeds 的方法.
另外在 ./jar目录里有flexlm.jar (10.1)版， 在./lib 目录里有liblmgr10.so 动态库 （Java Native C库）
我的问题是：
1. liblmgr10.so 是SDK 附带的文件还是经加用户key 编译的？是否有public key 的checksum隐含在其中？请问哪位大侠能提供一个flexlm 10.8 SDK 的Linux 版，谢谢！
2. 若liblmgr10.so文件是flexlm厂商提供的，我能否用任意的DWORD 和 已知的Java EncryptionSeeds异或生成一对Native C 的EncryptionSeeds?
3. 用Flexvkg3 生成的Native C 10.0 的VendorKey 和 已有的Java VendorKey 完全不一样：
/* Native C by Flexvkg3*/
/* Version 10 keys */
#define VENDOR_KEY1 0x131fd4e9
#define VENDOR_KEY2 0xc4ace148
#define VENDOR_KEY3 0x8632a4d2
#define VENDOR_KEY4 0x95e581be
#define VENDOR_KEY5 0x631a43c6
#define TRL_KEY1 0x20f258e2
#define TRL_KEY2 0x2d401ed5
// Java version
int[] EncryptionSeeds = { 0xe5fc8eb, 0x4b9ce6f0}; // 十进制 { 241158379, 1268573936 };
int[] VendorKeys = { 0x9b0ee7cb, 0x95512f20, 0x28b86f47, 0xf36d3518 }; // 十进制{ -1693521973, -1789841632, 683175751, -210946792 };
int[] CroKeys = { 0x600f78c5, 0x92422afd }; // 十进制{ 1611626693, -1841157379 };
能否是版本号是10.1的原因？ 哪位高人能搞一个能算10.1的算号器？
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